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AfilCa [17]
3 years ago
6

Write word equation for reaction between sodium and sulphur​

Chemistry
1 answer:
White raven [17]3 years ago
4 0

Answer:

2Na + S —> Na2S

Hope this helps.

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N₂(g) + 3H₂(g) → 2NH3(g)
Anna71 [15]

Answer:

93.5 moles N₂

Explanation:

To find the moles, you need to use the Ideal Gas Law. The equation looks like this:

PV = nRT

In this equation,

-----> P = pressure (atm)

-----> V = volume (L)

-----> n = number of moles

-----> R = constant (0.0821 atm*L/mol*K)

-----> T = temperature (K)

You can plug the given values into the equation and simplify to find moles. The final answer should have 3 sig figs to match the lowest number of sig figs among the given values.

P = 95.0 atm                  R = 0.0821 atm*L/mol*K

V = 224 L                       T = 2773 K

n = ?

PV = nRT

(95.0 atm)(224 L) = n(0.0821 atm*L/mol*K)(2773 K)

21280 = n(227.6633)

93.5 = n

8 0
2 years ago
How many grams of NaCl is needed to make 750mL of a 2.4 M NaCl solution?
Aleksandr [31]

Answer:

Molarity of the solution = 3.000 M

Volume of the solution = 250.0 mL = 0.25 L

moles in 250.0 mL = molarity x volume of the solution

                             = 3.000 M x 0.25 L

                             = 0.75 mol

Hence, 0.75 mol of NaCl is needed to prepare 250.0 mL of 3.000 M NaCl solution.

Moles (mol) = mass (g) / molar mass (g/mol)

Moles of NaCl in 250.0 mL = 0.75 mol

Molar mass of NaCl           = 58.44 g/mol

Mass of NaCl in 250.0 mL = Moles x Molar mass

                                         = 0.75 mol x 58.44 g/mol

                                         = 43.83 g

Hence, 43.83 g of NaCl is needed to prepare 250.0 mL of 3.000 M solution.

Explanation:

6 0
3 years ago
Protists are prokaryotes. t or f
fredd [130]
Protists are NOT prokaryotes so the answer would be B) False.
7 0
3 years ago
Which statement is true about the total mass of the reactants during a chemical change?
Mice21 [21]
It is greater than the total mass of the products
4 0
4 years ago
Enough of a monoprotic acid is dissolved in water to produce a 0.0121 M solution. The pH of the resulting solution is 6.49. Calc
Hitman42 [59]
Calculate the H positive from the pH equation: pH equals -log (H positive). This would be 10 to the -6.49. Let's call the acid HA. To calculate Ka in this equation, Ka equals H positive times A- over HA. HA is going to be the 0 0121. So, Ka=(10^-6.49)^2/0.0121. This equals 1.05*10^-13/0.0121. Ka then equals 8.65*10^-12.
4 0
3 years ago
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